Mathematics for the interested outsider

Differentiation Under the Integral Sign

Another question about “partial integrals” is when we can interchange integral and differential operations. Just like we found for continuity of partial integrals, this involves interchanging two limit processes.

Specifically, let’s again assume that is a function on the rectangle , and that is of bounded variation on and that the integral

exists for every . Further, assume that the partial derivative exists and is continuous throughout . Then the derivative of exists and is given by

Similar results hold where and are vector values, and where derivatives in terms of are replaced outside the integral by partial derivatives in terms of its components.

So, if and we can calculate the difference quotient:

where is some number between and that exists by the differential mean value theorem. Now to find the derivative, we take the limit of the difference quotient

as we take to approach , the number gets squeezed towards as well. Since we assumed to be continuous, the limit in the integrand will equal , as asserted.

Share this:

Like this:

Related

Doesn’t the differential MVT only imply that for each x there is a \xi(x) between y and y_0 such that…? It looks here that you’re applying the MVT to F(y) if we already know what F'(y) is, but I may be missing something.

As somebody trained in an engineering school, the temptation is to consider this obvious and skip right to the end without justification. It was enjoyable to see how the mean value theorem could be used to avoid such a handwaving argument.

Continuity of f(x,y) when y is varied appears to be required. I am guessing this is a requirement of the mean value theorem? Is there also a continuity requirement for x variation of f?

To extend this argument to the case where the boundaries are functions of y (, and ), I suppose you’d decompose the integral into sums. How to procede from there is not entirely clear to me.

John – I have a rather technical issue with your proof. You assume that is a function of x. It is not known to be a continuous function of x, and the mean value theorem only tells you that it exists, without giving you any properties of the function. Therefore, the integral isn’t even well defined by this argument (although, I’ve no doubt that will indeed be measurable).

I think you only need to show that $\latex (f(x,y)-f(x,y_0))/(y-y_0)\to D_2 (f(x,y_0)$ uniformly, which the MVT does prove, and this limit can be taken inside the integral.

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.